if x+y = 8 and x^2+y^2 = 34 then x*y will be

Rthy    11 years AGO

(x+y)^2 = 8^2
x^2 + y^2 + 2*x*y = 64
34 + 2*x*y =64 (Since x^2 + y^2 = 34)
2*x*y = 30
x*y = 15

Like? Yes (9) | No (2)  

souvik mondal   11 years AGO

x+y = 8
(x+y)^2=64
x^2+y^2+2xy=64
34+2xy=64
2xy=30
xy=15

Like? Yes (5) | No (1)  

A J V   11 years AGO

WKT, (x+y)^2 = (x^2)+(y^2)+(2*x*y)
Applying this eqn from the given values,
8^2=(34)+(2*x*y)
2*x*y = 64-34=> = 30
so, x*y= 15.
Hence, Option(8).

Like? Yes (2) | No   

anurag uttam   11 years AGO

x=5, y=3
x+y=8 then 5+3=8 and (5)^2+(3)^2=34
so x*y=15

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prem kumar.s   11 years AGO

(x+y)^2=x^2+y^2+2xy
therefore
8^2=34+2xy
64=34+2xy
30=2xy
15=xy
so ans is 15,and option 8th correct

Like? Yes (2) | No   

sainath   11 years AGO

(x+y)^2=x^2+y^2+2(a*b)
=>8^2=34+2(a*b)
=>64-34=2(a*b)
=>a*b=30/2
=>a*b=15

Like? Yes (2) | No   

Pranaya   11 years AGO

(x+y)^2=x^2+y^2+2*x*y=>8^2=34+2*x*y
x*y=30/2=15

Like? Yes (2) | No (1)  

Pallab Roy Chowdhury   11 years AGO

Given, x+y=8
=>y=8-x...(1)

and x^2+y^2=34
=>x^2+(8-x)^2=34[putting y=8-x]
=>x^2+{8^2-(2*8*x)+x^2}=34
=>x^2+64-16x+x^2=34
=>2x^2-16x+30=0
=>x^2-8x+15=0
=>x^2-5x-3x+15=0[Upon middle term factoring]
=>(x-5)(x-3)=0
Hence, x=5 or x=3.
When x=5 y=8-5=3 [From equation (1)]
and when x=3 y=8-3=5
In both cases x*y=5*3=3*5=15.

Like? Yes (3) | No (2)  

Narender Bhardwaj   11 years AGO

y= 8-x (1)
x^2+(8-x)^2=34
=>x^2+64+x^2-16x=34
=>(x-5)(x-3)=0
=>x=5,3.
from (1),y=3,5
Hence multiplication of x and y=x*y=15 ans

Like? Yes (2) | No (1)  

Balakumaran N.A   11 years AGO

lets consider formula: (x+y)^2=x^2+y^2+2xy
64=34+2xy
30=2xy then, xy =15.

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sagar arora   11 years AGO

x+y =8
so,(x+y)^2=64
And (x+Y)^2=x^2+y^2+2x*y
put the value of x^2+y^2=34
we get 2*x*y=30
or x*y=15

Like? Yes (1) | No   

ArunKumar   11 years AGO

(x+y)^2= x^2+y^2+2*x*y
x*y = [(x+y)^2- ( x^2+y^2)] / 2
x*y = [ 8^2 - 34 ] /2
= [64-34]/2
= 30/2
= 15

Like? Yes (1) | No   

raushan   11 years AGO

(x+y)^2=x^2+y^2+2*x*y
64=34+2*x*y
so xy=30/2
15

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Balakrishnan M   11 years AGO

(x+y)^2=x^2+y^2+2xy
8^2=34+2xy
64-34=2xy
30=2xy
xy=15

Like? Yes (1) | No   

Nikhil Mahapatra   11 years AGO

x + y = 8 --->(1)
x^2 + y^2 = 34---->(2)

from eq-(1) we can say:
(x + y)^2 = 8^2
x^2 + y^2 + 2xy = 64
from eq-(2) we know the value of x^2 + y^2 i.e 34

Hence,
34 + 2xy = 64
2xy = 64-34
xy = 30/2
xy = 15

So, Answer is 15 i.e option # 8

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parul mishra   11 years AGO

option(8)
using formula (x+y)^2= x^2+y^2+2*x*y
8^2=34+2*x*y
30=2xy
15=xy

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rambabu yalla   11 years AGO

(x+y)^2 =x^2+y^2-2xy
x*y=((x+y)^2-(x^2+y^2))/2
=(8^2-34)/2
=15

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J.P.Anitha   11 years AGO

=> (x+y)2 = x^2+y^2+2xy
=> 2xy=(x+y)^2 -(x^2+y^2)34
=> 2xy=8^2-34
=> 2xy=64-34
=> 2xy=30
=> x*y=30/2
=> Ans : 15 (Option :8)

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Keshav Gupta   11 years AGO

Assume that x=3 and y=5
so 3^2+5^2=34
then 3*5=15

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Bablu   11 years AGO

(x+y)^2=x^2+y^2+2xy
8^2=34+2xy
2xy=64-34=30
hence xy=30/2=15
Answer is (8) 15.

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jagannathan   11 years AGO

(x+y)^2=64
x^2+y^2+2xy=64
34+2xy=64
xy=15

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Lorna A. Canillo   11 years AGO

x+y=8 => y=8-x so, x^2+y^ => x^2(8-x)^2=34 then x^2+64-16x+x^2-34=0 => reduce to lowest term x^2-8x+15=0 therefore the values are 3 and 5. my answer is 15.

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Saurabh gupta   11 years AGO

ans= 15
(x+y)^2=64
then x^2+y^2+2xy=64
but x^2+y^2=34
hence 2xy=64-34
xy=15

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Vikrant Sahrawat   11 years AGO

x^2+y^2=(x+y)^2-2*x*y

So, 34=(x+y)^2-2*x*y

2xy=34-64
2xy=30
xy=15 (ans)

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swapnil gavtade   11 years AGO

(x+y)^2=64
x^2+2xy+y^2=64
x^2+y^2=34
2xy=30
xy=15

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ravi singh   11 years AGO

(x+y)^2=X^2+Y^2+2X*Y=64

put the value from equation therefore
x*Y=15

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Anurag   11 years AGO

nos. are 5 and 3 that we get on solving the equation

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Shailendra Agrawal   11 years AGO

3+5=8
3^2+5^2=34
3*5=15

Like? Yes (1) | No (1)  

susmit   11 years AGO

(x+y)^2= x^2+y^2+2xy
so 64=34+2xy
so 2xy=30
so xy=15

Like? Yes (1) | No (2)